I came across the integral $$ I = \int_{x_0}^{x_1} \frac{\mathrm{d}x}{\sqrt{x^3 + m x^2 + 1}} $$
I found the trivial solution involving elliptic integrals of the first kind, but I am wondering if there is a different way to solve it, perhaps involving hyperbolic functions, or something else?
Since $x_0, x_1$ are generic, we may as well consider the indefinite integral. For typical values of $m$ such an integral can't be expressed in terms of elementary functions, i.e., you need elliptic functions or the equivalent. As Claude Leibovici points out in the comments, if we express the cubic as $p(x) = (x - a) (x - b) (x - c)$, where $abc = -1$, $ab + bc + ca = 0$, $a + b + c = -m$, and $a < b < c$, we can express the integral more cleanly, as, e.g., $$\frac2{\sqrt{c - a}} F\left(\sqrt\frac{x - a}{b - a}, \sqrt\frac{b - a}{c - a}\right) + C ,$$ where $F$ is the incomplete elliptic integral function of the first kind.
For at least one real value of $m$ you can do better: If $m$ is the real root of the discriminant $-4m^2-27$ of the cubic polynomial $x^3 + m x^2 + 1$, the cubic polynomial has a multiple root $\alpha$ so the integral can be written as $$\int \frac{dx}{(x - \alpha) \sqrt{x - \beta}} = \frac{2}{\sqrt{\beta - \alpha}} \arctan \sqrt{\frac{x - \beta}{\beta - \alpha}} + C$$ where $\beta = -\frac\alpha2$ (in our case $\alpha < 0$).
